Surjection iff Right Cancellable/Necessary Condition/Proof 2

Theorem

Let $f$ be a surjection.

Then $f$ is right cancellable.

Let $f: X \to Y$ be surjective.

$\exists g: Y \to X: f \circ g = I_Y$

Suppose $h \circ f = k \circ f$ for two mappings $h: Y \to Z$ and $k: Y \to Z$.

Then:

$\ds h$

$=$

$\ds h \circ I_Y$

$\ds $

$=$

$\ds h \circ \paren {f \circ g}$

$\ds $

$=$

$\ds \paren {h \circ f} \circ g$

$\ds $

$=$

$\ds \paren {k \circ f} \circ g$

by hypothesis

$\ds $

$=$

$\ds k \circ \paren {f \circ g}$

$\ds $

$=$

$\ds k \circ I_Y$

$\ds $

$=$

$\ds k$

Thus $f$ is right cancellable.

So surjectivity implies right cancellability.

$\blacksquare$

1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.6$